Difference between revisions of "1999 CEMC Gauss (Grade 7) Problems/Problem 8"

Latest revision as of 10:49, 15 April 2025

Problem

The average of $10$ , $4$ , $8$ , $7$ , and $6$ is

$\text{(A)}\ 33 \qquad \text{(B)}\ 13 \qquad \text{(C)}\ 35 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 7$

Solution 1

The sum of all the listed numbers is

$10 + 4 + 8 + 7 + 6 = 35$

Since there are five numbers, we can divide the sum by $5$ to get the average:

$\frac{35}{5} = \boxed {\textbf {(E)} 7}$

Solution 2 (answer choices)

$10$ is too large of an answer because the largest number listed is $10$ but there are numbers that are less than $10$ , so the average will be lower than $10$ .

The only answer choice that is less than $10$ is $\boxed {\textbf {(E)} 7}$ .

Solution 3

From the list of numbers, we can observe that the numbers are all centered around 7. $8$ and $6$ are both $1$ away from $7$ , and $4$ and $10$ are both $3$ away from $7$ . $7$ is equal to itself.

Thus, the answer is $\boxed {\textbf {(E)} 7}$ .

@@ Line 2: / Line 2: @@
 The average of <math>10</math>, <math>4</math>, <math>8</math>, <math>7</math>, and <math>6</math> is
-<math>\text{(A)}\ 33  \qquad \text{(B)}\ 13  \qquad \text{(C)}\ 35 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ </math> 7
+<math>\text{(A)}\ 33  \qquad \text{(B)}\ 13  \qquad \text{(C)}\ 35 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 7</math>
 ==Solution 1==
 The sum of all the listed numbers is
@@ Line 15: / Line 15: @@
 The only answer choice that is less than <math>10</math> is <math>\boxed {\textbf {(E)} 7}</math>.
+==Solution 3==
+From the list of numbers, we can observe that the numbers are all centered around 7. <math>8</math> and <math>6</math> are both <math>1</math> away from <math>7</math>, and <math>4</math> and <math>10</math> are both <math>3</math> away from <math>7</math>. <math>7</math> is equal to itself.
+Thus, the answer is <math>\boxed {\textbf {(E)} 7}</math>.

Page

Toolbox

Search